The end B of the rod AB which makes angle the | vedclass

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Question

$V_{A} \cos (90-	heta)=V_{0} \cos 	heta$

$\mathrm{V}_{\mathrm{A}} \sin 	heta=\mathrm{V}_{0} \cos 	heta$

$\mathrm{V}_{\mathrm{A}}=\frac{\math

Vedclass · Accepted Answer

angular velocity of rod is $\frac{5}{3} \frac{v_0}{l}$

Vedclass · Answer

$V_{A} \cos (90-	heta)=V_{0} \cos 	heta$

$\mathrm{V}_{\mathrm{A}} \sin 	heta=\mathrm{V}_{0} \cos 	heta$

$\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{V}_{0}}{	an 	heta}$

$\mathrm{V}_{\mathrm{A}}=\frac{4}{3} \mathrm{V}_{0}$

$\omega=\frac{\left[V_{A} \sin (90-	heta)+V_{0} \sin 	hetaight]}{\ell}$

The end $B$ of the rod $AB$ which makes angle $\theta$ with the floor is being pulled with, a constant velocity $v_0$ as shown. The length of the rod is $l.$ At the instant when $\theta = 37^o $ then

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